Question

show that exactly one of the number N, N + 2 or 10 + 4 is divisible by 3

Answer 1

We applied Euclid Division algorithm on n and 3.
a = bq +r on putting a = n and b = 3
n = 3q +r , 0i.e n = 3q -------- (1),n = 3q +1 --------- (2), n = 3q +2 -----------(3)
n = 3q is divisible by 3
or n +2 = 3q +1+2 = 3q +3 also divisible by 3
or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3
Hence n, n+2 , n+4 are divisible by 3.

Answer 2

Checking the argument :- Only one of n, n+2 ,n+4 would be divisible by 3 when n is +ve integer.

Case 1 :-

n is divisible by 3 .

then n+2/3 = n/3 + 2/3 this shows a remainder 2 .

then n+4/3 =1+ n/3 + 1/3 this shows a remainder 1.

Case 2 :-

n+2 is divisible by 3 .

n/3 = n+2-2/3 = n+2/3 - 2/3 this leaves a remainder (3-2) = 1 .

n+4/3 =n+2+2/3 = n+2/3 + 2/3 leaves a remainder 2 .

Case 3 :-

n+4 is divisible by 3 .

n/3 = n+4-4/3 = (n+4)/3 -1 - 1/3 this leaves a remainder (3-1) = 2.

n+2/3 =n+4-2/3 = n+4/3 - 2/3 leaves a remainder (3-2)= 1 .

Hence proved that; one and only one out of n, n+2, n+4 is divisible by 3, where n is any positive integer